Advanced Calculus Single Variable

7.7 Exercises

= 0, is it necessary that x is either a local minimum or local maximum?
Hint:Consider f

(x)

= x3.

A continuous function f defined on

[a,b]

is to be maximized. It was shown above
in Theorem 7.6.2 that if the maximum value of f occurs at x ∈

(a,b)

, and if f is
differentiable there, then f′

(x)

= 0. However, this theorem does not say anything
about the case where the maximum of f occurs at either a or b. Describe how to
find the point of

[a,b]

where f achieves its maximum. Does f have a maximum?
Explain.

Show that if the maximum value of a function f differentiable on

[a,b]

occurs
at the right endpoint, then for all h > 0,f′

(b)

h ≥ 0. This is an example of
a variational inequality. Describe what happens if the maximum occurs at the
left end point and give a similar variational inequality. What is the situation for
minima?

Find the maximum and minimum values and the values of x where these are
achieved for the function f

(x)

= x +

√25-−-x2-

.

A piece of wire of length L is to be cut in two pieces. One piece is bent into
the shape of an equilateral triangle and the other piece is bent to form a square.
How should the wire be cut to maximize the sum of the areas of the two shapes?
How should the wire be bent to minimize the sum of the areas of the two shapes?
Hint:Be sure to consider the case where all the wire is devoted to one of the
shapes separately. This is a possible solution even though the derivative is not
zero there.

Lets find the point on the graph of y =

2
x4-

which is closest to

(0,1)

. One way to
do it is to observe that a typical point on the graph is of the form

( )
x, x42

and
then to minimize the function f

(x )

= x2 +

( x2 )
4-− 1

2. Taking the derivative of f
yields x +

14

x3 and setting this equal to 0 leads to the solution, x = 0. Therefore,
the point closest to

(0,1)

is

(0,0)

. Now lets do it another way. Lets use y =

x2
4

to
write x2 = 4y. Now for

(x,y)

on the graph, it follows it is of the form

(√4y, y)

.
Therefore, minimize f

(y)

= 4y +

(y− 1)

2. Take the derivative to obtain 2 + 2y
which requires y = −1. However, on this graph, y is never negative. What on earth
is the problem?

Find the dimensions of the largest rectangle that can be inscribed in the ellipse,

2
x9

+

2
y4-

= 1.

A function f, is said to be odd if f

(− x)

= −f

(x)

and a function is said to be
even if f

(− x)

= f

(x)

. Show that if f is even, then f′ is odd and if f is odd, then
f′ is even. Sketch the graph of a typical odd function and a typical even function.

Find the point on the curve, y =

√-------
25 − 2x

which is closest to

(0,0)

.

A street is 200 feet long and there are two lights located at the ends of the street.
One of the lights is

1
8

times as bright as the other. Assuming the brightness of
light from one of these street lights is proportional to the brightness of the light
and the reciprocal of the square of the distance from the light, locate the darkest
point on the street.

Find the volume of the smallest right circular cone which can be circumscribed
about a sphere of radius 4 inches.

PICT

Show that for r a rational number and y = xr, it must be the case that if this function
is differentiable, then y′ = rxr−1.

Let f be a continuous function defined on

[a,b]

. Let ε > 0 be given. Show there exists a
polynomial p such that for all x ∈

[a,b]

,

|f (x)− p(x)| < ε.

This follows from the Weierstrass approximation theorem, Theorem 6.10.3. Now here is
the interesting part. Show there exists a function g which is also continuous on

[a,b]

and
for all x ∈

[a,b]

,

|f (x)− g (x)| < ε

but g has no derivative at any point. Thus there are enough nowhere differentiable
functions that any continuous function is uniformly close to one. Explain
why every continuous function is the uniform limit of nowhere differentiable
functions. Also explain why every nowhere differentiable continuous function is the
uniform limit of polynomials. Hint: You should look at the construction of
the nowhere differentiable function which is everywhere continuous, given
above.

, etc. To go from Pn to Pn+1, delete the open interval which is the
middle third of each closed interval in Pn. Let P = ∩n=1∞Pn. By Problem 16 on
Page 155, P≠∅. If you have not worked this exercise, now is the time to do it.
Show the total length of intervals removed from

[0,1]

is equal to 1. If you feel
ambitious also show there is a one to one onto mapping of [0,1] to P. The set
P is called the Cantor set. Thus P has the same number of points in it as

[0,1]

in the sense that there is a one to one and onto mapping from one to
the other even though the length of the intervals removed equals 1. Hint:
There are various ways of doing this last part but the most enlightenment is
obtained by exploiting the construction of the Cantor set rather than some silly
representation in terms of sums of powers of two and three. All you need to do is use
the theorems in the chapter on set theory related to the Schroder Bernstein
theorem and show there is an onto map from the Cantor set to

[0,1]

. If you do
this right it will provide a construction which is very useful to prove some
even more surprising theorems which you may encounter later if you study
compact metric spaces. The Cantor set is just a simple version of what is seen
in some vegetables. Note in the following picture of a kind of broccoli, the
spirals of points each of which is a spiral of points each of which is a spiral of
points...

PIC

↑ Consider the sequence of functions defined in the following way. Let f1

(x)

= x on
[0,1]. To get from fn to fn+1, let fn+1 = fn on all intervals where fn is constant. If
fn is nonconstant on [a,b], let fn+1(a) = fn(a),fn+1(b) = fn(b),fn+1 is
piecewise linear and equal to

12

(fn(a) + fn(b)) on the middle third of [a,b].
Sketch a few of these and you will see the pattern. The process of modifying a
nonconstant section of the graph of this function is illustrated in the following
picture.

PICT

Show {fn} converges uniformly on [0,1]. If f(x) = limn→∞fn(x), show that
f(0) = 0,f(1) = 1,f is continuous, and f′(x) = 0 for all x

∕∈

P where P is the Cantor
set of Problem 14. This function is called the Cantor function.It is a very important
example to remember especially for those who like mathematical pathology. Note it has
derivative equal to zero on all those intervals which were removed and whose total
length was equal to 1 and yet it succeeds in climbing from 0 to 1. Isn’t this amazing?
Hint: This isn’t too hard if you focus on getting a careful estimate on the
difference between two successive functions in the list considering only a typical
small interval in which the change takes place. The above picture should be
helpful.